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Makovka662 [10]
3 years ago
10

An automotive manufacturer wants to know the proportion of new car buyers who prefer foreign cars over domestic. In an earlier s

tudy, the population proportion was estimated to be 0.33. How large a sample would be required in order to estimate the fraction of new car buyers who prefer foreign cars at the 90% confidence level with an error of at most 0.02?
Mathematics
1 answer:
anyanavicka [17]3 years ago
6 0

Answer:

1,496 new car buyers

Step-by-step explanation:

The sample size n in Simple Random Sampling is given by

\bf n=\frac{z^2p(1-p)}{e^2}

where  

z = 1.645 is the critical value for a 90% confidence level (*)

p= 0.33 is the population proportion.

e = 0.02 is the margin of error

so  

\bf n=\frac{(1.645)^20.33*0.67}{0.02^2}=1,495.76\approx 1,496

<em>(*)</em><em>This is a point z such that the area under the Normal curve N(0,1) inside the interval [-z, z] equals 90% = 0.9</em>

It can be obtained in Excel or OpenOffice Calc with

<em>NORMSINV(0.95)</em>

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